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Over [174]
3 years ago
10

Find the equation of the line that passes through the point left-parenthesis 1 comma 8 right-parenthesis and is perpendicular to

the line negative startfraction 4 over 5 endfraction x plus startfraction 5 over 4 endfraction y equals negative 4.
Mathematics
1 answer:
Furkat [3]3 years ago
7 0

Don't be shy about using actual parentheses and commas.

Line through (1,8) perpendicular to    

-\dfrac 4 5 x + \dfrac 5 4 y = -4

The perpendicular family is gotten by swapping the coefficients on x and y, negating exactly one of them.   The constant is given directly by the point we're through:

\dfrac 5 4 x + \dfrac 4 5 y = \dfrac 5 4 (1) + \dfrac 4 5 (8) = \dfrac 5 4 + \dfrac{32}{5}

Let's clear the fractions by multiplying both sides by 20.

25x + 16y = 25 + 128 = 153

Might as well stop here.

Answer: 25 x + 16 y = 153


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Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re
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Answer: R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}

Step-by-step explanation:

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\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}

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\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb

Resultant makes an angle of

\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}

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